The oxidation number of nickel in K4[Ni(CN)4] is:

Question

The oxidation number of nickel in K4[Ni(CN)4] is: (A) -2 (B) -1 (C) 2 (D) 0

Tardigrade Admin · Accepted Answer

Let the oxidation state of Ni be x in K 4[ Ni ( CN )4]. Since the overall charge on the complex is 0 , the sum of oxidation states of all elements in it should be equal to 0. Therefore, 4+x-4=0 text or, x=0

Q. The oxidation number of nickel in $K_{4} [N i (CN)_{4}]$ is:

-2

-1

2

0

Let the oxidation state of $N i$ be $x$ in $K_{4} [N i (CN)_{4}]$ . Since the overall charge on the complex is 0 , the sum of oxidation states of all elements in it should be equal to 0.
Therefore, $4 + x - 4 = 0$
$or, x = 0$

Q. The oxidation number of nickel in K4​[Ni(CN)4​] is:

-2

-1

2

0

Let the oxidation state of Ni be x in K4​[Ni(CN)4​]. Since the overall charge on the complex is 0 , the sum of oxidation states of all elements in it should be equal to 0. Therefore, 4+x−4=0 or, x=0

Q. The oxidation number of nickel in $K_{4} [N i (CN)_{4}]$ is:

Let the oxidation state of $N i$ be $x$ in $K_{4} [N i (CN)_{4}]$ . Since the overall charge on the complex is 0 , the sum of oxidation states of all elements in it should be equal to 0.
Therefore, $4 + x - 4 = 0$
$or, x = 0$